If a circuit has the current of (0.5t + 2.0i) amps and an impedance of (0.2 - 3.0i) ohms, find the voltage?

The impedance vector and current (Amps) vector of a given circuit must have the same phase angle.

Therefore:

t must have a value such that the absolute value of arc tan 2/.5t = the absolute value of arc tan -3/.2

l -3/.2 | must = | 2/.5t |

.5t = 2/15

t= 4/15

Amps = sqrt [(2/15)^2 + (2)^2] = 2.004439517 Amps

Power = (A^2)*(R) = (A^2(*(.2 Ohms) = .803555556 Watts

W = (V)*(A)*(pf)

Volts = (.803555556W)/ [(2.004439517A)*(cos arctan -3/.2)] = 6.027 Volts

If doing it by hand, convert to polar. The magnitude parts get divided, subract the angle part, convert back to rectangular. Otherwise, any decent engineering calculator can do complex division.

V=IR= (0.5t + 2.0i)(0.2-3.0i)

Use FOIL

V = 0.5t(0.2) +(0.5t)(-3.0i) + (2.0i)(0.2) +(2.0i)(-3.0i)

V = 1t -1.5it +0.4i -6.0i^2

but i^2 =-1

so V = t -1.5it +0.4i +6.0

grouping the like terms together

V = 6.0+t +(0.4- 1.5t)i